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8.3. A Brief Look at Some Advanced Ideas in Computer Vision 201
equivalent inhomogeneous formulation. Of course, there still is the little
matter of solving the linear system, but the DLT algorithm offers a solution
through singular value decomposition (SVD). For details of the SVD step,
which is a general and very useful technique in linear algebra, see [12].
Singular value decomposition factoriz es a matrix (which does not have to
be square) into the form A = LDV
T
,whereL and V are orthogonal matrices
(i.e., V
T
V = I) and D is a diagonal. (D is normally chosen to be square, but
it does not have to be.) So, for example, if A is 10 ×8thenL is 10 ×10, D is
10 ×8 and V
T
is 10 × 10. It is typical to arrange the elements in the matrix
D so that the largest diagonal coefficient is in the first row, the second largest
in the second row etc.
For example, the SVD of
⎡
⎣
11
01
10
⎤
⎦
=
⎡
⎢
⎣
√
6
3
0
√
3
3
√
6
6
−
√
6
6
−
√
3
3
√
6
6
√
2
2
−
√
3
3
⎤
⎥
⎦
⎡
⎣
√
30
01
10
⎤
⎦
√
2
2
√
2
2
√
2
2
−
√
2
2
.
There are two steps in the basic direct linear transform algorithm. They
are summarized in Figure 8.12. The DLT algorithm is a great starting point
for obtaining a transformation that will correct for a perspective projection,
but it also lies at the heart of more sophisticated estimation algorithms for
solving a number of other computer vision problems.
We have seen that the properties of any camera can be concisely specified
by the camera matrix. So we ask, is it possible to obtain the camera matrix
itself by examining correspondences from points (not necessarily lying on a
plane) in 3D space to points on the image plane? That is, we are examining
correspondences from P
3
to P
2
.
Of course the answer is yes. In the case we have just discussed, H was a
3 × 3 matrix whereas P is dimensioned 3 × 4. However, it is inevitable that
the estimation technique will have to be used to obtain the coefficients of P.
Furthermore, the only differences between the expressions for the coefficients
of H (Equation (8.6)) and those of P arise because the points in P
3
have four
coordinates, and as such we will have a m atrix of size 2n × 12. Thus the
analog of Equation (8.6) can be expressed as Ap = 0, and the requirement
for the estimation to work is n ≥ 6. (That is, 6 corresponding pairs with 12
coefficients and 11 degrees of freedom.)